Difference between revisions of "2017 AMC 8 Problems/Problem 25"
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==Solution== | ==Solution== | ||
− | Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>S'</math> and <math>T'</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>S'</math> and <math>T'</math>, and connect point <math>S'</math> with point <math>T'</math>. | + | Let the centers of the circles containing arcs <math>\overarc{SR}</math> and <math>\overarc{TR}</math> be <math>S'</math> and <math>T'</math>, respectively. Extend <math>\overline{US}</math> and <math>\overline{UT}</math> to <math>S'</math> and <math>T'</math>, and connect point <math>S'</math> with point <math>T'</math>. |
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<asy>draw((1,1.732)--(2,3.464)--(3,1.732)); | <asy>draw((1,1.732)--(2,3.464)--(3,1.732)); | ||
draw((1,1.732)--(0,0)--(4,0)--(3,1.732)); | draw((1,1.732)--(0,0)--(4,0)--(3,1.732)); | ||
Line 21: | Line 20: | ||
label("$S'$", (0,0), W); | label("$S'$", (0,0), W); | ||
label("$T'$", (4,0), E);</asy> | label("$T'$", (4,0), E);</asy> | ||
+ | We can clearly see that <math>\triangle AS'T'</math> is an equilateral triangle, because two of its angles are <math>60^\circ</math>, which is the degree measure of <math>\frac{1}{6}</math> a circle. The area of the figure is equal to the area of equilateral triangle <math>\triangle US'T'</math> minus the combined area of the <math>2</math> sectors of the circles. Using the area formula for an equilateral triangle, <math>\frac{\sqrt{3}}{4} \cdot a,</math> where <math>a</math> is the side length of the equilateral triangle, the area of <math>\triangle US'T'</math> is <math>\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.</math> The combined area of the <math>2</math> sectors is <math>2</math> times one sixth <math>\pi \cdot r^2</math>, which is <math>2 \cdot \frac 16 \cdot \pi \cdot 2^2 = \frac{4\pi}{3}.</math> Our final answer is then <math>\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.</math> | ||
==See Also== | ==See Also== |
Revision as of 00:42, 5 January 2018
Problem 25
In the figure shown, and are line segments each of length 2, and . Arcs and are each one-sixth of a circle with radius 2. What is the area of the region shown?
Solution
Let the centers of the circles containing arcs and be and , respectively. Extend and to and , and connect point with point . We can clearly see that is an equilateral triangle, because two of its angles are , which is the degree measure of a circle. The area of the figure is equal to the area of equilateral triangle minus the combined area of the sectors of the circles. Using the area formula for an equilateral triangle, where is the side length of the equilateral triangle, the area of is The combined area of the sectors is times one sixth , which is Our final answer is then
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
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All AJHSME/AMC 8 Problems and Solutions |
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